0
$\begingroup$

I'm working through an online course on Hypothesis Testing where a one-sample test of proportions is done using bootstrapping. I have some grounding in statistics, where the test statistic $ z $ for a one-sample test of proportions is mathematically calculated using te formula: $$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 * (1-p_0)}{n}}} $$, where $ p_0 $ is the hypothesized population proportion. The denominator is the standard error.

The course however estimates this using a bootstrap: it takes several resamples of a sample and takes the standard deviation of $ \hat{p} $ to be an estimate of the standard error, which can simply be mathematically calculated.

In related course, bootstrapping is used to estimate the standard error of the sampling distribution of the mean of a continuous variable. This standard error is then used to compute a confidence interval for the sample mean, using the CDF of a normal distribution.

My question is again similar to the proportions case: don't we already know the standard error of the sample mean? Isn't it just $ \frac{s}{\sqrt{n}} $ where $ s $ is the sample standard deviation and $ n $ the sample size? On top of this, we also know that the sampling distribution of the sample means follows a normal distribution (CLT) -- so we shouldn't need bootstrapping to generate a confidence interval.

Why do we need the bootstrap distribution to estimate standard error of the sampling distribution of a statistic?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ "we also know that the sampling distribution of the sample means follows a normal distribution (CLT)" The Central Limit Theorem makes a statement for sample size converging to infinity, and it has certain assumptions that are not always fulfilled. So at best we can approximate the sampling distribution of the mean with a normal distribution, and in a number of situations (small sample sizes, outliers, violation of i.i.d.) this approximation doesn't work well. $\endgroup$
    – Christian Hennig
    Commented May 8 at 9:10
  • 1
    $\begingroup$ Note that the bootstrap doesn't necessarily solve these issues, because it has trouble in similar situations as where the CLT is in doubt. $\endgroup$
    – Christian Hennig
    Commented May 8 at 9:17
  • 2
    $\begingroup$ "don't we already know the standard error of the sample mean? Isn't it just $\frac{s}{\sqrt{n}}$?" No. It is $\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the true unknown standard deviation, which we can estimate using $s$, so your formula gives an estimator of the standard error, which is often good but may not always be good. $\endgroup$
    – Christian Hennig
    Commented May 8 at 9:20

1 Answer 1

Reset to default
2
$\begingroup$

You don't need the bootstrap here. There's no real advantage to a simple bootstrap for the mean (though second-order bootstraps might be more accurate when the distribution of the mean is a bit further from Normal)

There are two arguments for using the bootstrap in this context, which are arguments for using the bootstrap basically everywhere that it applies.

The first is that the bootstrap works whether or not you have a simple analytic formula for the standard error of your statistic. You do for the mean, but often you don't. This means there's a case for using the bootstrap routinely, and only switching to other methods if the bootstrap is too slow or is not applicable.

The second argument is paedogogical. It turns out that resampling is quite a good way to teach the idea of sampling uncertainty in introductory classes (and that permutation tests are quite a good way to teach the idea of p-values, if you should wish to). One reference from New Zealand is here; we use resampling in high-school and first-year university statistics courses before we introduce sampling distributions like the Normal.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks - do you know of any sample statistic that fit this definition? Bootstrap works whether or not you have a simple analytic formula for the standard error of your statistic. You do for the mean, but often you don't $\endgroup$
    – WorldGov
    Commented May 8 at 8:31
  • 2
    $\begingroup$ The median. The interquartile range. The standard error of an odds ratio in logistic regression (there's an approximation here, but it's no better than the bootstrap) $\endgroup$
    – Thomas Lumley
    Commented May 8 at 8:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.